Exercise 1 (MC Questions)

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1 Dr. J. Tani Control Systems II (Spring 2018) Solution Exercise Set 3 Discrete-time Systems and Introduction to MIMO Systems Gioele Zardini, gzardini@ethz.ch, 12th March 2018 Exercise 1 (MC Questions) a) A =, B = A =, B = Solution. The eigenvalues of A must fulfill λ i < 1. (1) The first matrix has eigenvalues λ 1 = 0, λ 2 = 3. Because of the second one, the system is unstable. The second matrix has eigenvalues λ 1 = 0, λ 2 = 3. Because of the second one, the system is unstable. The third matrix has two 0 eigenvalues, which make the system asymptotically stable. The fourth matrix is an upper diagonal matrix and hence its eigenvalues can be seen on its diagonal, i.e. λ 1 = 1, λ 2 = 0.5. Because of the first one, the system is not asymptotically stable. The fifth/sixth matrix has igenvalues λ 1 = 0.1, λ 2 = 0.5, which make the system asymptotically stable. The seventh/eighth matrix has eigenvalues λ 1 = 0.1, λ 2 = 0.5, which make the system asymptotically stable. The ninth matrix has eigenvalues λ 1 = 0.1, λ 2 = 0.5, which make the system asymptotically stable. 1

2 b) True. False. Solution. True. In fact, the stability conditions only depend on matrix A. c) ( Z(s) = P12 (s)k(s)(i P 22 (s)k(s)) 1 P 21 (s) + P 11 (s) ) W (s). Z(s) = ( P 22 (s)k(s)(i P 12 (s)k(s)) 1 P 21 (s) + P 11 (s) ) W (s). Z(s) = ( P 22 (s)k(s)(i P 12 (s)k(s)) 1 P 11 (s) + P 21 (s) ) W (s). Z(s) = ( P 12 (s)k(s)(i P 21 (s)k(s)) 1 P 12 (s) + P 11 (s) ) W (s). P 22 + P 21 + u K P 12 w + + P 11 z Figure 1: MIMO loop. Solution. By introducing a dummy signal u(t), one can write Z(s) = P 12 (s)k(s)u(s) + P 11 (s)w (s). (2) In order to find U(s) let s analyze the sum it is subject to: U(s) = P 22 (s)k(s)u(s) + P 21 (s)w (s) (I P 22 (s)k(s))u(s) = P 21 (s)w (s) U(s) = (I P 22 (s)k(s)) 1 P 21 (s)w (s). (3) By plugging Equation (3) into Equation (2), one gets Z(s) = P 12 (s)k(s)u(s) + P 11 (s)w (s) = P 12 (s)k(s)(i P 22 (s)k(s)) 1 P 21 (s)w (s) + P 11 (s)w (s) ( ) = P 12 (s)k(s)(i P 22 (s)k(s)) 1 P 21 (s) + P 11 (s) W (s). (4) d) Solution. Since matrix A multiplies the state space vector x(t) and there are four states, in order for the multiplication to be feasible it must have four columns. Since matrix A is a square matrix, it must have four rows as well. 2

3 Matrix Number of rows Number of columns A 4 4 B 4 3 C 2 4 D 2 3 Table 1: Matrix dimensions. Since matrix B multiplies the input vector u(t) and there are three inputs, in order for the multiplication to be feasible it must have three columns. The vector resulting from the multiplication should be added to Ax(t) R 4 1. Since the sum between matrices is only possible for matrices with the same dimension, matrix B should have four rows. Since matrix C multiplies the state space vector and the multiplication must result in a vector 2-by-1 vector (there are two outputs), in order for the multiplication to be feasible, C R 2 4, i.e. must have 2 rows 4 columns. Since matrix D multiplies the input vector, it must have 3 columns. Since the result should be a 2-by-1 vector (two outputs), matrix D must have 2 rows. 3

4 Exercise 2 (Continuous-time and discrete-time stability) a) For C(s) to be asymptotically stable, its pole π c = α must lie in the left half of the complex plane: Re{π c } < 0 α > 0. (5) b) The Euler backward emulation approach leads to the substitution s = z 1 T z. If one substitutes this into the transfer function of the continuous-time controller, one gets (6) C(z) = 2 z 1 T z + 1 z 1 T z + α = z (2 + T ) 2 z (1 + α T ) 1. (7) c) The controller C(z) is asymptotically stable if its poles π d,i fulfill the condition π d,i < 1. (8) One can read the pole from Equation (7): z (1 + α T ) 1 = 0 1 π d = 1 + α T. (9) This, together with the condition for stability, gives 1 < α T < 1 α > 0 or α < 2 T. (10) Together with the results from (b), the condition on α for both systems to be asymptotically stable is α > 0. 4

5 Exercise 3 (Difference Equations) You are given a controller transfer function C(s) = 2(s + 0.5). s + 1 (11) a) Starting from the definition of the controller transfer function, i.e. C(s) = U(s) E(s), (12) one can write U(s)(1 + s) = (2s + 1)E(s) U(s) + su(s) = 2sE(s) + E(s). (13) Using the differential rule for Laplace transform and applying the inverse Laplace transform to both sides of the equation one gets u(t) + u(t) = e(t) + 2ė(t). (14) b) Using Euler backward method, the differential equation becomes u[k] u[k 1] 2(e[k] e[k 1]) u[k] + = e[k] + T s T s u[k] (1 + 1 ) u[k 1] 2(e[k] e[k 1]) = + e[k] + Ts T s T s u[k 1] T u[k] = s + e[k] + 2(e[k] e[k 1]) ( T ) s Ts (15) = u[k 1] + (T s + 2)e[k] 2e[k 1]. (T s + 1) 5

6 Exercise 4 a) First order minors are: (Poles and Zeros of MIMO Systems: Minors Method) 1 s + 1, s 1 (s + 1)(s + 2), 1 s 1, 1 s + 2, 1 s + 2 (16) Minors of second order are: (s 1) (s + 1)(s + 2) 2, 2 (s + 1)(s + 2), 1 (s + 1)(s + 2) (17) The least common denominator the pole-polynom is (s + 1)(s + 2) 2 (s 1), from which the system s poles and their multiplicities can be read: s = 1 (multiplicity = 1), s = 1 (multiplicity = 1) und s = 2 (multiplicity = 2). b) Normalizing the minors of second order with the denominator (s + 1)(s + 2) 2 (s 1) yields (s 1) 2 (s + 1)(s + 2) 2 (s 1), 2(s 1)(s + 2) (s + 1)(s + 2) 2 (s 1), (s 1)(s + 2) (s + 1)(s + 2) 2 (s 1) (18) The greatest common divisor of these minors is (s 1) and therefore the MIMO system has its only zero at s = 1. c) In a), four poles were found. Therefore, the system has an internal description of at least fourth order. Comment: A MIMO system can have poles and zeros at the same frequency without a pole zero cancellation taking place. In order to have a pole zero cancellation, the pole and zero have to be at the same frequency and in the same direction. 6

7 Exercise 5 (Push-Through Rule) It holds G 1 G 1 G 2 G 1 = G 1 (I G 2 G 1 ) = (I G 1 G 2 )G 1. (19) Multiplying both sides of the equation from the left by (I G 1 G 2 ) 1 one gets (I G 1 G 2 ) 1 G 1 (I G 2 G 1 ) = G 1. (20) Multiplying both sides of the equation from the right by (I G 2 G 1 ) 1 one gets (I G 1 G 2 ) 1 G 1 = G 1 (I G 2 G 1 ) 1. (21) 7

8 Exercise 6 (MIMO Sensitivity Functions) We present here two different solutions, which differ by the starting point of the derivation. Starting from the error E(s) If one wants to determine the matrices of the sensitivity transfer functions, one can start writing (by paying attention to the direction of multiplications) E(s) and Y (s). By looking at the given control loop, one writes E(s) = R(s) P (s) C(s) E(s) D(s), Y (s) = P (s) C(s) E(s) + D(s). (22) This gives in the first place E(s) = (I + P (s) C(s)) 1 (R(s) D(s)). (23) Plugging this into the expression for Y (s) and dropping the s in the transfer functions for simplicity (i.e. F (s) = F ), results in Y = P C E + D = P C (I + P C) 1 (R D) + D = P C (I + P C) 1 R P C (I + P C) 1 D + (I + P C) (I + P C) 1 D }{{} I = P C (I + P C) 1 R + (I + P C P C) (I + P C) 1 D = P C (I + P C) 1 R + (I + P C) 1 D. (24) Recalling the general relation one gets Y (s) = T (s) R(s) + S(s) D(s), (25) T 1 (s) = P (s) C(s) (I + P (s) C(s)) 1, S 1 (s) = (I + P (s) C(s)) 1. (26) Starting from the input U(s) If one wants to determine the matrices of the sensitivity transfer functions, one can start writing (by paying attention to the direction of multiplications) U(s) and Y (s). By looking at the given control loop, one writes U(s) = C(s) (R(s) D(s)) C(s) P (s) U(s), Y (s) = P (s) U(s) + D(s). (27) This gives in the first place U(s) = (I + C(s) P (s)) 1 C(s) (R(s) D(s)). (28) Plugging this into the expression for Y (s) and dropping the s in the transfer functions for simplicity (i.e. F (s) = F ), results in Y = P U + D = P (I + C P ) 1 C (R D) + D ( ) = P (I + C P ) 1 C R + I P (I + C P ) 1 C D. (29) 8

9 Recalling the general relation one gets Y (s) = T (s) R(s) + S(s) D(s), (30) T 2 (s) = P (s) (I + C(s) P (s)) 1 C(s), S 2 (s) = I P (s) (I + C(s) P (s)) 1 C(s). (31) It can be shown that this two different results actually are the equivalent. It holds S 1 = S 2 (I + P C) 1 = I P (I + C P ) 1 C I = I + P C P (I + C P ) 1 C (I + P C) I = I + P C P (I + C P ) 1 (C + C P C) I = I + P C P (I + C P ) 1 (I + C P ) C I = I + P C P C I = I (32) T 1 = T 2 P C (I + P C) 1 = P (I + C P ) 1 C P C = P (I + C P ) 1 C (I + P C) P C = P (I + C P ) 1 (C + C P C) P C = P (I + C P ) 1 (I + C P ) C P C = C P I = I. Finally, one can show that S(s) + T (s) = (I + P C) 1 + P C (I + P C) 1 = (I + P C) (I + P C) 1 = I. (33) 9

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